Question

sinx=3/5,x lies in second quadrant find all other trigonometric ratios​

Answer 1

Step-by-step explanation:

1 - Sin^2x = Cos ^2x

Cos^2x = 1 - (3/5)^2

Cos^2x = 1 - 9/25

Cos^2x = 16/25

Cos x = - 4/5 {2nd quadrant}

Cosec x = 5/3

Sec x = - 5/4

Tan x = - 3/4

Cot x = - 4/3

Answer 2

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

$\tt{\rightarrow sinx=-\dfrac{3}{5}}$

As we know that :-

$\tt{\rightarrow cosecx=\dfrac{1}{sinx}}$

$\tt{\rightarrow cosecx=\dfrac{1}{(3/5)}}$

$\tt{\rightarrow cosecx=\dfrac{5}{3}}$

Also we know that :-

${\boxed{\sf\:{sin^2x+cos^2x=1}}}$

Hence,

cos²x = 1 - sin²x

${\boxed{\sf\:{Putting\;the\;values\;:-}}}$

$\tt{\rightarrow cos^2x=1-(\dfrac{3}{5})^2}$

$\tt{\rightarrow cos^2x=1-\dfrac{9}{25}}$

$\tt{\rightarrow cos^2x=\dfrac{16}{25}}$

Therefore,

$\tt{\rightarrow cosx=\pm\dfrac{4}{5}}$

Here we get,

x lies in second quadrant.

Hence,

Value of cosx will be negative.

Now,

$\tt{\rightarrow cosx=-\dfrac{4}{5}}$

$\tt{\rightarrow secx=\dfrac{1}{cosx}}$

$\tt{\rightarrow secx=\dfrac{1}{(-4/5)}}$

$\tt{\rightarrow secx=-\dfrac{5}{4}}$

Now,

$\tt{\rightarrow tanx=\dfrac{sinx}{cosx}}$

$\tt{\rightarrow tanx=\dfrac{(3/5)}{-(4/5)}}$

$\tt{\rightarrow tanx=-\dfrac{3}{4}}$

Now,

$\tt{\rightarrow cotx=\dfrac{1}{tanx}}$

$\tt{\rightarrow cotx=-\dfrac{4}{3}}$